(2p+1)^2 +(p+1)=p
When:
p: prime number
My questions
mathematics questions
1- What conditions does this equation need to fulfill in order to become a mathematical conjecture? The guess is that there are an infinite number of prime numbers written on this picture
If someone can tell me the percent...
This can not be true. It is $(2p+1)^2>0$, so the LHS can never equal the RHS. Edit: Is your question which prime numbers $p$ satisfy that $(2p+1)^2+p$ is an other prime $q$?
Well, the sum of two primes is always even (except one prime is 2). So it kind off can never hold (except one prime number is assumed to be 2). Do you mean different primes in this equation? $(2a+1)^2+b=c$ where $a,b,c$ are prime? Else this will always fail.
As I said, in that case you always have to assume that $b=2$. Else you get "odd + odd = even" so there is kind off no chance.
Sorry, you are absolutely right. The mistake was from me. I forgot to write part of the equation, so it turned out like that. This is the equation I was telling you about
I will give you examples. (11*2+1)^2+12=541 prime number
Well, I think I can not help you with this question, as you have asked for other examples. I can not provide a programm. But certainly this will be a tough question to answer, as most questions about such identities are pretty hard to solve.
I think this is a valid conjecture to ask if there are an infinite amount of primes satisfying this equation. But it looks at least as difficult to the twin prime conjecture.
Discussion between Mohammed and Cornman
Discussion between Mohammed and Cornman